Mean Value Theorem: Definition, Theorem, Proof and Explanation

Ans.In 1823, Augustin Louis Cauchy came up with the modern form of the Mean Value Theorem. Initially it was Michel Rolle who in the year 1691 came up with the Rolle Theorem which is the restricted form of the Mean Value theorem which was applicable to only polynomials without including the techniques of calculus. 

Ans. The Mean Value Theorem is the extension of The Rolle’s Theorem. While the Mean Value Theorem states that let f be the continuous function on closed interval [a,b] and differentiable on open interval (a,b), where a

Ans. In the year 1691, A french mathematician Michel Rolle created The Rolle’s theorem, Also the theorem as we can see, is indeed named after him.

Ans. We have, f(x) = |x| =  { -x, when – 1 ≤ x ≤ 0 and x, when 0 ≤ x ≤ 1}

As polynominal function is everywhere continuous and at the same time differentiable, therefore f(x) is continuous and differentiable for all x < 0 and for all x> 0 except possibly at x = 0.

Thus, consider the point x = 0.

We have, lim f(x) = lim – x = 0 and lim f(x) = lim x = 0 

x → 0 ^–     x → 0 ^–                x → 0⁺       x → 0 ^– 

∴ lim f(x) = lim f(x) = f(0) 

    x → 0 ^–    x → 0⁺

Thus, f(x) is continuous at x = 0

Hence, f(x) is continuous on [-1, 1]

Now, (LHD at x = 0) = lim \(\frac{f(x) - f(0)}{x - 0}\)

                                x → 0 ^– 

(LHD at x = 0) = lim \(\frac{- x - 0}{x}\) = – 1  [Q f(x) = – x for x< 0 and f(0) = 0]

                                 x → 0 

and, (RHD at x = 0) = lim \(\frac{f(x) - f(0)}{x - 0}\)= lim \(\frac{x - 0}{x}\) = 1

                                  x → 0⁺                                  x → 0 ^–

[Q f(x) = x for x³ 0]

\ (LHD at x = o) ¹ RHD at x = 0

This shows thar f(x) is not differentiable at x = 0 \(\hat{I}\) ( – 1, 1).

Thus, the condition of derivability at each point of ( – 1, 1) is not satisfied.

Hence, Rolle’s theorem is not applicable to f(x) = |x| on [ – 1, 1]

Ans. (i) Here, the function is polynomial which is continuous and differentiable everywhere and so will be continuous on [ – 1, 3] and differentiable on ( – 1, 3)

A couple of quick function evaluations show that f ( – 1) = f (3) = – 5.

Thus, the conditions for Rolle’s theorem is met and so we can solve the problem by taking the derivative.

f’ (x) = 2x – 2

and solve, f’ (c)= 0

2c – 2 = 0    ⇒  c = 1.

We found a single value and it is in the interval so the value we want is, c = 1.

(ii) Even here, the function is polynomial which is continuous and differentiable everywhere and so will be continuous on [ – 2, 1] and differentiable on ( – 2, 1).

A couple of quick function evaluations show that g ( – 2) = g (1) = 0

The conditions for Rolle’s theorem is met and so we can solve the problem by taking the derivative.

g’ (t) = 2 – 2t – 3t²

and then by solving g’ (c) = 0,

 – 3c² – 2c + 2 = 0    ⇒  c = \(\frac{ 1 \pm \sqrt{7}} {- 3} = - 1.2153, 0.5486\)

In this case, they are both in the interval so the values are,

c = \(\frac{ 1 \pm \sqrt{7}} {- 3} = - 1.2153, 0.5486\)

Ans. Given, the function f(x) = x² ( x – 1) = x³ – x² and the interval [ 0, 3]. This function is continuous on the closed interval [ 0, 3] as it is polynominal. And the derivative of ‘f’ is f’(x) = 3x² – 2x.

We can now observe that f is differentiable on the open interval (0,3). The assumptions of the mean value theorem have now been met. Let’s now apply the mean value theorem and find out all possible values of c in the open interval (0, 3). Then,

f’ (c) = \(\frac{f (3) - f (0)}{3 - 0}\)→

3c² – 2c = \(\frac{((3)^3 - (3)^2)) - ((0)^3 - (0)^2)}{3 - 0}\)→

3c² – 2c = \(\frac{18 - 0}{3}\)→

3c² – 2c = 6 →

3c² – 2c – 6 = 0 →

(3)c² + ( – 2)c + ( – 6) = 0 →

Now, by using the quadratic formula,

c = \(\frac{- (-2) \pm \sqrt{(-2)^2 - 4(3)(- 6)}}{2(3)}\)→

c = \(\frac{2 \pm \sqrt{76}}{6}\)→

c = \(\frac{2 \pm 2 \sqrt{19}}{6}\)→

c = \(\frac{1 \pm \sqrt{19}}{3}\)→

c = \(\frac{1 \pm \sqrt{19}}{3}\)≈ 1.786 or \(\frac{1 - \sqrt{19}}{3}\)≈ 1.120 ( – 1.120 is NOT in the intercal (0,3).

c =\(\frac{1 \pm \sqrt{19}}{3}\)≈ 1.786

Ans. We are given the function \(f (x) = \frac{x}{1+x}\) and the interval [ 1, 3]. This function is continuous on the closed interval [ 1, 3] as it is the quotient of continuous functions y = x (polynominal) and y = 1 +x (polynominal). And the derivative of f is, 

\(f' (x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}\)

We can now observe that f is differentiable on the open interval (1, 3). The assumptions of the mean value theorem have now been met. Let’s now apply the mean value theorem and find out all possible values of c in the open interval (1, 3). Then,

f’ (c) = \(\frac{f(3) - f(1)}{3-1}\)→

\(\frac{1}{(1+c)^2} = \frac{ \frac{(3)}{1+(3)} - \frac{(1)}{(1+(1)}}{3-1}\)→

\(\frac{1}{(1+c)^2} = \frac{ \frac{(3)}{4} - \frac{(1)}{2}}{ \frac{2}{1}}\)→

\(\frac{1}{(1+c)^2} = (\frac{3}{4} - \frac{2}{4} ). { \frac{1}{2}}\)→

\(\frac{1}{(1+c)^2} = \frac{1}{8}\)→

(1 + c) ² = 8 →

1 + c = \(\pm\) \(\sqrt{8}\) = \(\pm\) 2 \(\sqrt{2}\) →

c = \(\pm\)2\(\sqrt{2}\) – 1 →

c = 2 \(\sqrt{2}\) – 1 ≈ 1.828 or – 2 \(\sqrt{2}\) – 1 ≈ 3.828 (3.828 is NOT in the interval (1,3).

c = 2 \(\sqrt{2}\) – 1 ≈ 1.828 →

Ans. In order to find the values, we must find the derivates first.

g(x) = x³ + x² – x

g’(x) = 3x² + 2x – 1

Then we must figure the slope between the endpoints of the interval.

g ( – 2) = ( – 2)³ + ( – 2)² – ( – 2)

= – 2 

g(1) = 1

m = \(\frac{g (-2) - g(1)}{-2-1}\)

= \(\frac{-2-1}{-2-1}\)

= 1

Finally, by setting the derivative equal to this slope,

3x² + 2x – 1 = 1

3x² + 2x – 2 = 0

x = \(\frac{ -2 \pm \sqrt{4 - (-24)}}{6}\)

= \(\frac{ -2 \pm 2 \sqrt{7}}{6}\)

= \(\frac{ - 1 - \sqrt{7}}{3} \) or \(\frac{ - 1 + \sqrt{7}}{3} \)

Therefore, the values are \(\frac{ - 1 - \sqrt{7}}{3} \) and \(\frac{ - 1 + \sqrt{7}}{3} \)where both are inside the given interval.